Two positive charges of magnitude q are place | vedclass

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Question

Initial potential of the charge,

$V_{A}=\frac{2 k q}{a}-\frac{2 k q}{a \sqrt{5}}$

$\Rightarrow V_{A}=\frac{1}{4 \pi E} \frac{2 q}{a}\left(1-\fra

Vedclass · Accepted Answer

$\frac{1}{{4\pi \,\,{\varepsilon_{0}}}}\,\frac{{2qQ}}{a}\,\left( {1\,\, - \,\frac{1}{{\sqrt 5 }}} \right)$

Vedclass · Answer

Initial potential of the charge,

$V_{A}=\frac{2 k q}{a}-\frac{2 k q}{a \sqrt{5}}$

$\Rightarrow V_{A}=\frac{1}{4 \pi E} \frac{2 q}{a}\left(1-\frac{1}{\sqrt{5}}ight)$

(Here potential due to each $q=\frac{k q}{a}$ and potential due to each $-q=\frac{-k q}{a \sqrt{5}}$)

Final potential of the charge

$V_{B}=0$

( $\because$ Point $B$ is equidistant from all the four charges)

$	herefore$ Using work energy theorem,

$\left(W_{A B}ight)_{	ext {electric }}=Q\left(V_{A}-V_{B}ight)$

$=\frac{2 q Q}{4 \pi E_{0} a}\left[1-\frac{1}{\sqrt{5}}ight]$

$=\left(\frac{1}{4 \pi \varepsilon_{0}}ight) \frac{2 Q q}{a}\left[1-\frac{1}{\sqrt{5}}ight]$

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